# Prove that 1+1=2 [duplicate]

Possible Duplicate:
How do I convince someone that $1+1=2$ may not necessarily be true?

I once read that some mathematicians provided a very length proof of $1+1=2$.

Can you think of some way to extend mathematical rigor to present a long proof of that equation? I'm not asking for a proof, but rather for some outline what one would consider to make the derivation as long as possible.

EDIT: It seems the proof I heard about is a standard reference given here multiple times :) I stated that the proof itself is less useful than an outline for me as I know only "physics level maths". Can someone provide a short outline what's going on in the proof? Some outline I can look up section by section in Wikipedia to at least get a feel of what could be needed to make such a proof?

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## marked as duplicate by Antonio Vargas, Sasha, Austin Mohr, Did, Jason Bourne Jan 15 '13 at 1:02

Usually, it is defined to be that $2 := 1 + 1$, $3 := 1 + 1 + 1$, etc. What definition of $2$ are you using? –  andybenji Jan 15 '13 at 0:23
What is there to prove? Thats how we define the integer two, its doesn't require a proof because its a definition. –  Ethan Jan 15 '13 at 0:24
Here's a page from Russel and Whitehead's Principia Mathematica proving that 1+1=2, on page 379. quod.lib.umich.edu/cgi/t/text/… –  user7530 Jan 15 '13 at 0:25
Have you read these two answers (math.stackexchange.com/questions/95069/… and math.stackexchange.com/questions/243049/…) of mine? –  Asaf Karagila Jan 15 '13 at 0:31
Note that the point of the work of Russell and Whitehead wasn't to make the derivation as long as possible. The point was to put the foundation of mathematics on firmly rigorous ground in a highly axiomatized system. The "long proof" is more accurately a book that builds up this system before eventually proving that $1+1=2$. –  Code-Guru Jan 15 '13 at 0:45

You are thinking of the Principia Mathematica, written by Alfred North Whitehead and Bertrand Russell. Here is a relevant excerpt:

As you can see, it ends with "From this proposition it will follow, when arithmetical addition has been defined, that 1+1=2." The theorem above, $\ast54\cdot43$, is already a couple of hundred pages into the book (Wikipedia says 370 or so); the later theorem alluded to, that $1+1=2$, appears in section $\ast102$, considerably farther on.

I wrote a blog article a few years ago that discusses this in some detail. You may want to skip the stuff at the beginning about the historical context of Principia Mathematica. But the main point of the article is to explain the theorem above.

The article explains the idiosyncratic and mostly obsolete notation that Principia Mathematica uses, and how the proof works. The theorem here is essentially that if $\alpha$ and $\beta$ are disjoint sets with exactly one element each, then their union has exactly two elements. This is established based on very slightly simpler theorems, for example that if $\alpha$ is the set that contains $x$ and nothing else, and $\beta$ is the set that contains $y$ and nothing else, then $\alpha \cup \beta$ contains two elements if and only if $x\ne y$.

The main reason that it takes so long to get to $1+1=2$ is that Principia Mathematica starts from almost nothing, and works its way up in very tiny, incremental steps. The work of G. Peano shows that it's not hard to produce a useful set of axioms that can prove 1+1=2 much more easily than Whitehead and Russell do.

It would not be hard to copy-and-paste the relevant parts of the blog article here, but I am not sure if that is appropriate se.math etiquette; I invite comments on this matter.

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Thanks. I have to check that. Looks like one of the most useful answers here :) –  Gerenuk Jan 15 '13 at 0:56
See Arithmetic of Peano axioms in Wikipedia. Set $0+1=1$ and use (in)formal recursion for definition of $$\begin{array}{rrll} +:&\mathbb{N}\times\mathbb{N}&\longrightarrow &\mathbb{N}\\ & (n,\varphi(m)) &\longmapsto & \varphi \big( n+m\big) \end{array}$$ if $m\neq 0$. Here the bijection $\varphi:\mathbb{N}\to\mathbb{N}\big\backslash\{0\}$ is the sucessor function: $$\varphi(0)=1\\ \varphi(1)=2\\ \vdots\\ \varphi(n)=n+1.$$
Your use of "$\varphi^{-1}$" is rather suspicious here, since it isn't well-defined. Usually, one says that $+$ takes $(n, 0)$ to $n$, and $(n, \varphi(m))$ to $\varphi(+(n, m))$. Your definition neglects to define $1+0$. –  MJD Jan 15 '13 at 1:02